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binomial theorem

find and simplify the term independent of x in the expansion of (1/2x + x^3)^8

do you do this the same as if you were to find the coefficient of X^2 but find the number without the x?

thanks

Reply 1

spread_logic_not_hate

teenagedirtbag

find and simplify the term independent of x in the expansion of (1/2x + x^3)^8
do you do this the same as if you were to find the coefficient of X^2 but find the number without the x?

thanks

Yeah you're looking for the term without x in it, i.e. the term with x^0 in it.

Reply 2

teenagedirtbag

spread_logic_not_hate

Yeah you're looking for the term without x in it, i.e. the term with x^0 in it.

how do i do that though, because i tried, but both terms have x in and i've got really confused.

thanks

Reply 3

musabjilani

It's the one with (1/2x)^6 and (x^3)^2. You just need to look for the term in which the x's cancel.

Reply 4

teenagedirtbag

musabjilani

It's the one with (1/2x)^6 and (x^3)^2. You just need to look for the term in which the x's cancel.

i got to that stage, but i dont seem to be able to get the x's to cancel. unless im just doing something completely wrong :s-smilie:

Reply 5

rollerball900

Just to clarify:
You cant understand how to get the power of zero?

Reply 6

spread_logic_not_hate

teenagedirtbag

i got to that stage, but i dont seem to be able to get the x's to cancel. unless im just doing something completely wrong :s-smilie:

So you've got to find

(\dfrac{1}{2x})^6 \times (x^3)^2 = \dfrac{1}{2^6} \times \dfrac{1}{x^6} \times x^6 = \dfrac{1}{2^6}

That's the terms that will be without x. Now all you gotta do is multiply it by the right binomial coefficient for the answer, which will be 8 C 6 in this case.